$\begin{aligned} g(x)&=\sqrt{\sin(x)} \\\\ g'(x)&=\,? \end{aligned}$ Choose 1 answer: Choose 1 answer: (Choice A) A $\dfrac{[\sin(x)]^{-\frac{1}{2}}}{2}$ (Choice B) B $\dfrac{\cos(x)}{2\sqrt{\sin(x)}}$ (Choice C) C $\sqrt{\cos(x)}$ (Choice D) D $\dfrac{\cos(\sqrt{x})}{2\sqrt{x}}$
Since $g$ is a composite function, we can use the chain rule. The chain rule says: $\dfrac{d}{dx}\left[w\Bigl(u(x)\Bigr)\right]={w'\Bigl({u(x)}\Bigr)}\cdot{u'(x)}$ To use it, we need to recognize our function as a composite function like $w\big(u(x)\big)$. $g(x)=\underbrace{\sqrt{~\overbrace{\sin(x)}^{\text{inner}}~}}_{\text{outer}}$ So if $g(x)=w(u(x))$, then: $\begin{aligned} {u(x)}&={\sin(x)} &&\text{inner function} \\\\ w(x)&=\sqrt{x}&&\text{outer function} \end{aligned}$ Before applying the chain rule, let's find the derivatives of the inner and outer functions (work not shown, but hopefully these derivatives are familiar by now). $\begin{aligned} {u'(x)}&={\cos(x)} \\\\ {w'(x)}&={\dfrac{1}{2\sqrt{x}}} \end{aligned}$ Now let's apply the chain rule: $\begin{aligned} g'(x)&=\dfrac{d}{dx}\left[w\Bigl(u(x)\Bigr)\right] \\\\ &={w'\Bigl({u(x)}\Bigr)}\cdot{u'(x)} \\\\ &={\dfrac{1}{2\sqrt{{\sin(x)}}}} \cdot {\cos(x)} \\\\ &=\dfrac{\cos(x)}{2\sqrt{\sin(x)}} \end{aligned}$